How do you multiply #sqrt(-10)*sqrt(-40)#?

2 Answers
Jun 13, 2015

Answer:

-20

Explanation:

When working in the complex numbers, we need to remember that the rule #sqrta sqrtb = sqrt(ab)# only works if #a# and #b# are not both negative.

We can change #sqrt(-10) = isqrt(10)#.

And then use

#sqrt(-10)sqrt (-40) = isqrt10sqrt(-40) = isqrt (-400) = i(20i) = -20#.

Or we can also rewrite #sqrt(-40) = isqrt40#, so we have:

#sqrt(-10) sqrt(-40) = isqrt(10) i sqrt(40) = i^2 sqrt 400 = -20#.

Jun 13, 2015

Answer:

#sqrt(-10)*sqrt(-40) = +- 20#

Explanation:

#sqrt(-10) * sqrt(-40) = (+-i sqrt(10)) * (+-i sqrt(40))#

#= +-i^2 sqrt(10)*sqrt(40) = +-sqrt(10*40) = +-sqrt(400) = +-20#

The problem here is that #sqrt(-10)# and #sqrt(-40)# are not uniquely defined since #sqrt(-1)# is not uniquely defined.

If #a in RR# and #a > 0# then #sqrt(a)# denotes the positive square root of #a#. It has another square root, viz #-sqrt(a)#.

if #a < 0# then #a# has two pure imaginary square roots which you could call #+-i sqrt(-a)#.

From the perspective of #RR#, the number #i# which we call the square root of #-1# is indistiguishable from #-i#. We cannot pick one of #i# or #-i# as #sqrt(-1)# by saying we want the positive one.